We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle.
In any triangle the sides of a triangle are proportional to the sines of the angles opposite to them.
That is in any triangle ABC,
a/sin A = b/sin B = c/sin c
Proof:
Let ABC be a triangle.
Now will derive the three different cases:
Case I: Acute angled triangle (three angles are acute): The triangle ABC is acuteangled.
Now, draw AD from A which is perpendicular to BC. Clearly, D lies on BC
Now from the triangle ABD, we have,
sin B = AD/AB
⇒ sin B = AD/c, [Since, AB = c]
⇒ AD= c sin B ……………………………………. (1)
Again from the triangle ACD we have,
sin C = AD/AC
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ...………………………………….. (2)
Now, from (1) and (2) we get,
c sin B = b sin C
⇒ b/sin B = c/sin c………………………………….(3)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = c/sin c………………………………….(4)
Therefore, from (3) and (4) we get,
a/sin A = b/sin B = c/sin C
Case II: Obtuse angled triangle (one angle is obtuse): The triangle ABC is obtuse angled.
Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.
Now from the triangle ABD, we have,
sin ∠ABD = AD/AB
⇒ sin (180  B) = AD/c, [Since ∠ABD = 180  B and AB = c]
⇒ sin B = AD/c, [Since sin (180  θ) = sin θ]
⇒ AD = c sin B ……………………………………. (5)
Again, from the triangle ACD, we have,
sin C = AD/AC
⇒ sin C = AD/b, [Since, AC = b]
⇒ AD = b sin C ……………………………………. (6)
Now, from (5) and (6) we get,
c sin B = b sin C
b/sin B = c/sin C ……………………………………. (7)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = b/sin B ……………………………………. (8)
Therefore, from (7) and (8) we get,
a/sin A = b/sin B = c/sin C
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle C is a right angle.
Now from triangle ABC, we have,
sin C = sin π/2
⇒ sin C = 1, [Since, sin π/2 = 1], ……………………………………. (9)
sin A = BC/AB
⇒ sin A = a/c, [Since, BC = a and AB = c]
⇒ c = a/sin A ……………………………………. (10)
and sin B = AC/AB
⇒ sin B = b/c, [Since, AC = b and AB = c]
⇒ c = b/sin B ……………………………………. (11)
Now from (10) and (11) we get,
a/sin A = b/sin B = c
⇒ a/sin A = b/sin B = c/1
Now from (9) we get,
⇒ a/sin A = b/sin B = c/sin C
Therefore, from all three cases, we get,
a/sin A = b/sin B = c/sin C. Proved.
Note:
1. The sine rule or the law of sines can be expressed as
sin A/a = sin B/b = sin C/c
2. The sine rule or the law of sines is a very useful rule to express sides of a triangle in terms of the sines of angles and viceversa in the following manner.
We have a/sin A = b/sin B = c/sin C = k\(_{1}\) (say)
⇒ a = k\(_{1}\) sin A, b = k\(_{1}\) sin B and c = k\(_{1}\) sin C
Similarly, sin A/a = sin B/b = sin C/c = k\(_{2}\) (say)
⇒ sin A = k\(_{2}\) a, sin B = k\(_{2}\) b and sin C = k\(_{2}\) c
Solved problem using the law of sines:
The triangle ABC is isosceles; if ∠A = 108°, find the value of a : b.
Solution:
Since the triangle ABC is isosceles and A = 108°, A + B + C = 180°, hence it is evident that B = C.
Now, B + C = 180°  A = 180°  108°
⇒ 2B = 72° [Since, C = B]
⇒ B = 36°
Again, we have, \(\frac{a}{sin A}\) = \(\frac{b}{sin B}\)
Therefore, \(\frac{a}{b}\) = \(\frac{sin A}{sin B}\) = \(\frac{sin 108°}{sin 36°}\) = \(\frac{cos 18°}{sin 36°}\)
Now, cos 18° = \(\sqrt{1  sin^{2} 18°}\)
= \(\sqrt{1  (\frac{\sqrt{5}  1}{4})^{2}}\)
= ¼\(\sqrt{10 + 2\sqrt{5}}\)
and sin 36° = \(\sqrt{1  cos^{2} 36°}\)
= \(\sqrt{1  (\frac{\sqrt{5} + 1}{4})^{2}}\)
= ¼\(\sqrt{10  2\sqrt{5}}\)
Therefore, a/b = \(\frac{\frac{1}{4}\sqrt{10 + 2\sqrt{5}}}{\frac{1}{4}\sqrt{10  2\sqrt{5}}}\)
= \(\frac{\sqrt{10 + 2\sqrt{5}}}{\sqrt{10  2\sqrt{5}}}\)
= \(\sqrt{\frac{(10 + 2\sqrt{5})^{2}}{10^{2}  (2\sqrt{5})^{2}}}\)
= \(\frac{10 + 2\sqrt{5}}{\sqrt{80}}\)
⇒ \(\frac{a}{b}\) = \(\frac{2√5(√5 + 1)}{4 √5}\)
⇒ \(\frac{a}{b}\) = \(\frac{√5 + 1}{2}\)
Therefore, a : b = (√5 + 1) : 2
11 and 12 Grade Math
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